So, where's the error...

nfeese	Posted 10/24/2005 12:21 PM (#221) Subject: So, where's the error...
Math Location: IAKSS	MathCounts Puzzle of the Week: If x=.9 repeating, then 10x = 9.9 repeating. So, 10x - x = 9.9 repeating -.9 repeating. If that is true, then, 9x/9 = 9/9 so, x must = 1 and therefore .9 repeating actually equals 1. We know that is not true and therefore the problem is incorrect. So, where's the error?
bperry	Posted 10/24/2005 4:26 PM (#225 - in reply to #221) Subject: RE: So, where's the error...
Math Posts: 281 Location: Henry Clay	.9 repeating does equal 1
bperry	Posted 10/24/2005 4:27 PM (#226 - in reply to #221) Subject: RE: So, where's the error...
Math Posts: 281 Location: Henry Clay	1/3 = .3 repeating 1=3(1/3) = 3(.3 repeating) = .9 repeating
rmcquerr	Posted 10/24/2005 9:09 PM (#228 - in reply to #221) Subject: RE: So, where's the error...
Math Posts: 79 Location: Tates Creek	.9 repeating is actually one of the only few ambiguous cases in mathematics. Its in an error in the notation we have accepted for decimal expansions. Another proof that .9 repeating equals one is through geometric series. 9\10 + 9\100 + 9\1000 + ... So a = 9\10 and r = 1\10, thus a/(1-r) = 1.
bperry	Posted 12/17/2005 11:59 PM (#468 - in reply to #221) Subject: RE: So, where's the error...
Math Posts: 281 Location: Henry Clay	Thought this was funny: (series.jpg) Attachments ---------------- series.jpg (20KB - 436 downloads)